Answer :
Answer
The projectile reaches maximum height at t = 2.5 s.
And the highest point the projectile will reach is s = 100 ft.
Explanation
The quation for the projectile fired is
s(t) = 80t - 16t²
with s(t) the height in feet and t is the time in seconds.
We are then asked to find the time that the projectile will reach its highest point and how high it will be.
We already have a function of how the projectile's height varies with time, so, finding information about the highest point the projectile reaches will just be like finding the maximum for that height function.
s(t) = 80t - 16t²
At the maximum point for any function, the first derivative is 0, and the second derivative is usually negative or less than 0.
s(t) = 80t - 16t²
First derivative will have us differentiate the function with respect to t.
(ds/dt) = 80 - 32t
Recall that at maximum point, first derivative = 0
(ds/dt) = 80 - 32t = 0
80 - 32t = 0
-32t = -80
Divide both sides by -32
(-32t/-32) = (-80/-32)
t = 2.5 s
To confirm that this time corresponds to the maximum point of the function, we take the second derivative
Second derivative will have us differentiate the first derivative with respect to t
(ds/dt) = 80 - 32t
(d²s/dt²) = 0 - 32 = -32 < 0
Hence, t = 2.5 s is the time when the projectile reaches maximum point.
To now obtain the maximum height, we put t = 2.5 s into the function for the height.
s(t) = 80t - 16t²
At maximum height, t = 2.5 s
s(t=2.5) = 80(2.5) - 16(2.5²)
= 200 - 100
= 100 ft.
Hope this Helps!!!
